Elevation of Boiling Point

Elevation of Boiling Point:

The temperature of a substance (liquid) at which the vapour pressure of a liquid is equal to atmospheric pressure is called the boiling point of the liquid. It is characteristic of each liquid. For example- the boiling point of H₂O is 373 K.

We know that the vapour pressure of a solution containing a non-volatile solute is lower than the vapour pressure of the pure solvent. Thus, such solutions have to be heated more to boil the solution.

This increase in temperature or the additional temperature required to boil such solution is called elevation in boiling point. If T°_b is the boiling point of the solvent and T_b is the boiling point of the solution, then, ΔT_b is the elevation in boiling point which can be mathematically written as-

ΔT_b = (T_b – T°_b)

This elevation in boiling point is colligative property as the elevation in boiling point is directly proportional to the lowering of vapour pressure (ΔP),

i.e., ΔT_b ∝ ΔP ……….(i)

But according to Raoult’s Law,
ΔP ∝ x_B ……….(ii) [Where x_b is the mole fraction of the solute]

From (i) and (ii), ΔT_b ∝ x_B
⇒ ΔT_b = k . x_B [Where k is the constant of proportionality]
⇒ ΔT_b = (k . n_B) / (n_A + n_B) = k (n_B/n_A)
Since solution is dilute and n_B << n_A [as ΔT_b is colligative property]
⇒ ΔT_b = k . n_B / (W_A/M_A) = k . M_An_B / W_A ……….(iii)
[Where W_A = Weight of solvent and M_A = Molecular Mass of solvent]

If the weight of solvent W_A = 1 kg; then n_B / W_A = molality (m) of the solution.

Therefore, ΔT_b = k . M_A . m = k_b . m ……….(iv)
[Where k_b = k . M_A is another constant]

This new constant k_b is called molal elevation constant or ebullioscopic constant.

If m = 1 ⇒ ΔT_b = k_b

Hence, the molal elevation constant is the elevation in boiling point when the molality of the solution is unity. This constant has a characteristic value for each liquid which changes with temperature. Its unit is K/m or K kg mol^-1.

Example- For H₂O, k_b = 0.52 K kg mol^-1 at 373 K.

Calculation of Molecular Mass of Solute from ΔT_b:

From relation (iv), ΔT_b = k_b . m = k_b (n_B / W_A) ……….(v)
Since molality is the number of moles of solute in 1000 gms of solvent,
Therefore, m = (W_B/M_B) X (1000/W_A) ……….(vi)
Using relation (vi) in (v), we get
ΔT_b = (k_b . W_B . 1000) / (M_B . W_A)
Therefore, M_B = (k_b . W_B . 1000) / (ΔT_b . W_A)
Hence molar masses of non-volatile solute are determined.

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